1. ## Collatz Conjecture Investigation

Hi, I am currently investigating the Collatz conjecture out of curiosity, having watched a video made about it by Numberphile.

My current understanding of the problem is that any odd $n$, written as $2^{a_s}(2b_s-1)-1$, will iterate to even $3^{a_s}(2a_s-1)-1$ after $a_s$ iterations of $\frac{3n+1}{2}$, then being divided by 2 until it becomes some odd $2^{a_{s+1}}(2b_{s+1}-1)-1$.

When $3^a\left(2b-1\right)-1$ is written as $3^ab-\left(3^a+1\right)$, factoring out the highest power of 2 shows that for $a \equiv b \pmod 2$, the highest power of 2 dividing $3^a\left(2b-1\right)-1$ is 2, and for $a \equiv 1 \pmod 2$ where $b \equiv 0 \pmod 4$, or $a \equiv 0 \pmod 2$ where $b \equiv 3 \pmod 4$, it must be 4. In other cases it seems to be no longer dependant on $a$ or $b$ and becomes pseudo-random.

If the Collatz conjecture is true, for any combination of $(a_1,b_1)$ there must be some integer $z$ such that both $a_z$ and $b_z$ equal 1 after $z$ of iterations of $2^{a_s}(2b_s-1)-1 \rightarrow 2^{a_{s+1}}(2b_{s+1}-1)-1$.

An expansion of the $z$th iteration of $2^{a_s}(2b_s-1)-1 \rightarrow 2^{a_{s+1}}(2b_{s+1}-1)-1$ on some odd $n=2^{a_1}(2b_1-1)-1$, $T_z(n_1)$, implies that:
$$T_z(n_1)=\frac{1.5^{\sum_{c=2}^{z}a_c}\left(1.5^ {a_1}\left(n_1+1\right)-1\right)}{2^{\sum_{d=1}^{z}b_d}}+\frac{1.5^{a_z}-1}{2^{b_z}}+\sum_{e=2}^{z-1}\frac{1.5^{\sum_{f=e+1}^{z}a_f}\left(1.5^{a_e}-1\right)}{2^{\sum_{g=e}^{z}b_g}}=1$$
must have a unique solution for any $n_1$ such that $\left\{z,a_{1...z},b_{1...z}\in\mathbb{N^+}\right \}$

I am trying to find a connection between $(a_s,b_s)$ and $(a_{s+1},b_{s+1})$, such that an expression for the sum of these values for each iteration may be found, which would make the above expression more useful. Corrections and advice on whether this may or may not be possible would really be appreciated!

Thanks

2. ## Re: Collatz Conjecture Investigation

There is a reason that some of the greatest mathematical minds of our generation have declared that mathematics is not yet ready for this problem. You will need to create some new math. There is no uniform relation between $(a_s,b_s)$ and $(a_{s+1},b_{s+1})$. One investigation that I saw tried to find patterns for them. But, in general, they can be any natural number.

3. ## Re: Collatz Conjecture Investigation

Also note that if you take the binary representation of $2^{a_s}(2b_s-1)-1$, the number of ending 1 is the number of time $(a_s)$ you can apply $\frac{3n+1}{2}$ before reaching $3^{a_s}(2b_s-1)-1$.

4. ## Re: Collatz Conjecture Investigation

There should be some kind of connection but it is not easy to make a formula out of it.
If $2b_s-1$ is of the form $8x+5$ and $a_s$ is even, $3^{a_s}(2b_s-1)-1$ is divisible exactly by $2^2$ to produce the next odd. If $a_s$ is odd, it is divisible exactly by $2^1$.
If it is of the form $8x+7$ it is the opposite ($a_s$ even -> divisible by $2^1$, $a_s$ odd -> divisible by $2^2$,).
Other forms gives other exponents, but they seems regular....with some logic behind.
But even knowing it is divisible by $2^2$ does not give your $a_{s+1}$ straight away.